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$f:\Bbb R \rightarrow (2,+\infty)$

and it is known that $$f^3(x)-12f(x)=3x-15$$ I need to verify f's monotony but without the use of the derivative so

Let there be a second function $$H(x)=x^3-12x \qquad \forall x \epsilon (2,+\infty)$$ so we got this $$H (f (x))=3x-15$$$$\forall x_1,x_2 \epsilon \Bbb R : x_1 \lt x_2 \Rightarrow 3x_1-15 \lt 3x_2-15 \Rightarrow H (f (x_1)) \lt H (f (x_2))$$ so I need to verify the monotony of $H $ but I cant do it without derivative.

Antonis Sk
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1 Answers1

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We have

$$H(z)-H(y)=(z^3-12z)-(y^3-12y)=(z^3-y^3)-12(z-y)$$

Now

$$z^3-y^3=(z-y)(z^2+yz+y^2)$$

so

$$H(z)-H(y)=(z-y)[(z^2+yz+y^2)-12]>0$$

if $z>y>2$

smcc
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  • can you explain why z^2 +yz+y^2 is bigger than 12 ??? when z and y is bigger than 2? – Antonis Sk Aug 19 '16 at 15:54
  • When $z=y=2$ we have $z^2+yz+y^2=4+4+4=12$, and when $y,z>2$, the expression $z^2+yz+y^2$ is increasing because $z^2$ and $yz$ are increasing in $z$ (if you do not want to take that as given, you can prove it fairly easily). – smcc Aug 19 '16 at 17:04
  • alright i got it ... thanks again – Antonis Sk Aug 19 '16 at 17:18
  • I am actually a bit unsure why you wanted to do it without the derivative. We can differentiate $H$ without making any assumptions about the differentiability of $f$ – smcc Aug 19 '16 at 17:43
  • well this is an excersise of a friend who doesnt know how to use derivatives – Antonis Sk Aug 19 '16 at 17:46